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java - 弹出 ListView 出现在另一个 ListView 的 ItemClickListener 上

转载 作者:行者123 更新时间:2023-11-29 23:27:17 25 4
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我的应用程序中有一个 ListView,当我单击第一个 ListView 的每个项目时,我想要一个弹出窗口 ListView 出现,但是必须在弹出 View 中的第二个 ListView 不会出现。这是我的整个 OnCreate 代码:

@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
ListView users = findViewById(R.id.users);
String[] values = new String[] { "A",
"B",
"C",
"D",
"E"
};
ArrayAdapter<String> adapter = new ArrayAdapter<String>(this,
android.R.layout.simple_list_item_1, android.R.id.text1, values);
users.setAdapter(adapter);
users.setOnItemClickListener(new AdapterView.OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
ListView todo =new ListView(MainActivity.this);
String[] itemss={"Kick","Make Admin","Mute"};
ArrayAdapter<String> adapter=new ArrayAdapter<String>(view.getContext(),
android.R.layout.simple_list_item_1,android.R.id.text1,itemss);
todo.setAdapter(adapter);
todo.setOnItemClickListener(new AdapterView.OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> parent, View view, int
position, long id) {
ViewGroup vg=(ViewGroup)view;
TextView txt=(TextView)vg.findViewById(R.id.txtitem);
Toast.makeText(MainActivity.this,txt.getText().toString(),Toast.LENGTH_LONG).show();
final CharSequence[] items = {"Delete User","Add User","Ban User"};
final ArrayList selectedItems=new ArrayList();
AlertDialog.Builder builder = new AlertDialog.Builder(view.getContext());
builder.setTitle("Admin accesses");
builder.setMultiChoiceItems(items, null, new DialogInterface.OnMultiChoiceClickListener() {
@Override
public void onClick(DialogInterface dialog, int which, boolean isChecked) {
if (isChecked) {
selectedItems.add(which);
} else if (selectedItems.contains(which)) {
selectedItems.remove(Integer.valueOf(which));
}
}
});
builder.setPositiveButton("DONE", new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
}
});
builder.setNegativeButton("CANCEL", new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
dialog.dismiss();
}
});
builder.create();
builder.show();
}
});
}
});
}

正如您在代码中看到的那样,我希望通过单击每个第二个 ListView 项来弹出复选框,但首先我必须解决出现第二个 ListView 的问题有谁知道我该如何解决这个问题?这是否可能在另一个 ListViewOnItemVlickListener 上显示一个 ListView

最佳答案

没有添加第二个 ListView 到父 View ( Activity/您的父布局,如相对布局/线性布局等)。

You need to add your listview todo to the parent layout and need to define layout params.

例如:-

    RelativeLayout relativeLayoutParent = (RelativeLayout) findViewById(R.id.your_relative_layout_parent);
ListView todo=new ListView(this);
ViewGroup.LayoutParams layoutParams=new ViewGroup.LayoutParams(ViewGroup.LayoutParams.MATCH_PARENT,ViewGroup.LayoutParams.MATCH_PARENT);
todo.setLayoutParams(layoutParams);
relativeLayoutParent.addView(todo);

In above example, I make assumption that your layout parent is Relative layout.

对于弹出窗口:-

将下面的代码更改为

 builder.create();
builder.show();

这个

AlertDialog dialog=builder.create();
dialog.show();

它将显示您的多个 chicle 警报对话框。

关于java - 弹出 ListView 出现在另一个 ListView 的 ItemClickListener 上,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53350670/

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